What is the derivative of (2x/(x^2-1)) ?

The derivative of (2x/(x^2-1)) is (2(-x^2-1))/((x^2-1)^2)

What is the first derivative of (2x/(x^2-1)) ?

The first derivative of (2x/(x^2-1)) is (2(-x^2-1))/((x^2-1)^2)

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\frac{d}{d x} (\frac{2 x}{x ^{2} - 1})

\frac{2 ( - x ^{2} - 1 )}{( x ^{2} - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x}{x ^{2} - 1})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (\frac{x}{x ^{2} - 1})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 2 \cdot \frac{\frac{d x}{d x} ( x ^{2} - 1 ) - \frac{d}{d x} ( x ^{2} - 1 ) x}{( x ^{2} - 1 ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (x^{2} - 1) = 2 x

= 2 \cdot \frac{1 \cdot ( x ^{2} - 1 ) - 2 xx}{( x ^{2} - 1 ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot ( x ^{2} - 1 ) - 2 xx}{( x ^{2} - 1 ) ^{2}} : \frac{2 ( - x ^{2} - 1 )}{( x ^{2} - 1 ) ^{2}}

= \frac{2 ( - x ^{2} - 1 )}{( x ^{2} - 1 ) ^{2}}

x \to 0 - lim (- x)

f (x) = e^{1 - x}

\frac{\partial}{\partial x} (6 e^{x y + 3})

n = 0 \sum \infty \frac{1}{n ^{\frac{1}{2}}}

\frac{x ^{2}}{3 ( 2 - x )}

What is the derivative of (2x/(x^2-1)) ?
The derivative of (2x/(x^2-1)) is (2(-x^2-1))/((x^2-1)^2)
What is the first derivative of (2x/(x^2-1)) ?
The first derivative of (2x/(x^2-1)) is (2(-x^2-1))/((x^2-1)^2)